Question

A student carried heated a 25.00 g piece of aluminum to a temperature of 100°C, and placed it in 100.00 g of water, initially at a temperature of 10.0°C. Determine the final temperature of the system (aluminum and water)
cH2OB4.18J/gc

c Aluminum .900 j/g c

Answer 1

Answer: The final temperature of the system is 14.60°C

Explanation:

When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed by water.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$      ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of aluminium = 25.00 g

$m_2$ = mass of water = 100 g

$T_{final}$ = final temperature = ?°C

$T_1$ = initial temperature of aluminium = 100°C

$T_2$ = initial temperature of water = 10°C

$c_1$ = specific heat of aluminium = 0.900 J/g°C

$c_2$ = specific heat of water= 4.18 J/g°C

Putting values in equation 1, we get:

$25\times 0.900\times (T_{final}-100)=-[100\times 4.18\times (T_{final}-10)]$

$T_{final}=14.60^oC$

Hence, the final temperature of the system is 14.60°C